Download Conclusion Proof ADB= CDB=900 AD=DC AB=BC DAB= DCB

Conclusion
⦟ADB=⦟CDB=900
AD=DC
AB=BC
⦟DAB=⦟DCB
⦟DCB+⦟DBC=90o
⦟ABD+⦟DAB=90o
⦟ABD+⦟DAB=⦟DCB+⦟DBC=90o
⦟ABD=⦟DBC
⦟DBC=⦟ABD=(⦟ABC)/2
Proof
Line BD is a perpendicular bisector of line AC
Line BD is a perpendicular bisector of line AC
Applying pythagoras theorem to the conclusions
reached above
Given AB=BC
Given that all the angles in a triangle add up to
180o and ⦟CDB=900
Given that all the angles in a triangle add up to
180o and ⦟ADB=900
Given ⦟ADB=⦟CDB=900 and that all angles
within a triangle add up to 180o
Substituting ⦟DAB with ⦟DCB in
⦟ABD+⦟DAB=⦟DCB+⦟DBC and subtracting
⦟DCB from both sides of the resulting equation,
the equation in the right hand column remains
The sum of ⦟DBC and ⦟ABD is ⦟ABC and since
they are equal, each of them is half their sum
Line BD bisects ⦟ABC
The two angles on either side of BD are equal
and their sum is ⦟ABC
Conclusion
Sin(⦟DBC)=sin(⦟ABD)
Proof
Given AD=DC and applying the sine rule to both
triangles
Given that ⦟ADB=⦟CDB=900 and that all the
angles within a triangle add up to 90o
Only equal acute angles give equal sines.
The two angles ⦟DBC and ⦟ABD are equal and
they add up to ⦟ABC therefore each of them is
half of ⦟ABC
The angles ⦟DBC and ⦟ABD are on either side of
the line and are equal
⦟DBC and ⦟ABD are acute angles
⦟ABD=⦟DBC
⦟DBC=⦟ABD=(⦟ABC)/2
Line BD is a bisector of ⦟ABC